What's with Sableye?

Yes but they are not independant variables. What you said would be assuming you would be considering that drawing a card would have no effect on the next card that you drew.


I know 98% sounds absurd but it is true. Its much like the birthday problem that sounds crazy. If there are 25 people in the same room there is roughly a 50% chance that 2 people will share the same birthday. It sounds crazy but it is true. If you need me to explain that I will too.
 
52 Cards. Alright that is a given. Yet to simplify this wouldn't it make more sense to water this down to a deck of cards? So take a deck of playing cards. Then set aside 1 of them (remember that you set aside one for the Wager.) Shuffle your current hand into your deck and draw 6 cards. What are your chances of grabbing and Ace and a King? Less than 98%? I thought so.

Also,you don't get to evolve until next turn anyways. So wouldn't it just make more sense to play a viable draw supporter and bolster your chances?
 
So after you draw that first card the percent goes up around 1.5%, and goes up from there, but it still doesn't make it 98% of the time. I know they are not independent variables, but you understand what I'm getting at. Each card draw depends on the card you drew before, but it is so much simpler to just count from the starting point rather then compute each varied draw. Also, don't forget what Benjamin Disraeli said, "There are three kinds of lies: lies, damned lies, and statistics.". So skew whatever you want from information you want, it still a matter of interpretation.
 
you would be surprised DOX I am in a college statistics course and when I first started I assumed these things. I dont mean to sound like Im being condescending at all, but You have to use combinations and permutations to solve these problems. Otherwise you are solving for an independent case which is when a first draw does not effect the second draw. If you are in college and high school approach your math or statistics professor about this. It blew me away when I learned how to do this.

I dont feel like getting on your bad sides so I am just going to drop it.
 
I do see where you are comming from. (Honor Math Student) Yet I still feel that your Math isn't 100% correct. Since in your original map out of the whole thing, you forgot about the discarded Wager. That would mean that the chances would go up. You can't go too much further than 98%, not only that but you could also have attached an energy, also raising the chances. So what if you attached and energy and accounted for the TGW? Would you hit 99%? I still find the whole thing ridiculous.
 
Technically he's talking about standard deviations and outermost variables lying outside the normal plot so 98% is actually correct, but I don't think it applies to cards effectively even if it is stats.
 
ericmeckley said:
you would be surprised DOX I am in a college statistics course and when I first started I assumed these things. I don't mean to sound like Im being condescending at all, but You have to use combinations and permutations to solve these problems. Otherwise you are solving for an independent case which is when a first draw does not effect the second draw. If you are in college and high school approach your math or statistics professor about this. It blew me away when I learned how to do this.

I don't feel like getting on your bad sides so I am just going to drop it.
Click to expand...

You surely have a flaw in your argument. I'm no Stats College person, but just an AP Calc guy. However...
ericmeckley said:
say you have your 6 prizes and 2 basics out. That leaves 52 cards in your deck. Assuming that you have no candies or Stage 2 donks prized that leaves 4 of each in your 52 card deck. If you win the Wager you would draw six cards. To find the probability of drawing 1 of each you would use the formula...4C1 x 4C1 x 44C4 / 52C6 = about .98. Now I will explain the formula... The first 4C1 represents 4 rare candies in deck and choosing one of them. The second 4C1 represents 4 stage two donks in the deck choosing 1 of them. The 44C4 represents the remainder of the cards in the deck and choosing 4 of them. All of this gets divided by 52C6 which is your deck choosing 6 cards.
Click to expand...

Let's argue from the opposite direction. What are the chances of not getting a rare candy/Stage 2? Out of 52 cards, you have 8 undesired cards. That's 44. You draw 6 cards:
(44/52)(43/51)(42/50)(41/49)(40/48)(39/47). Evaluate and you get 34.6737% of not getting a rare candy or a Stage 2. Argue that, because I don't see a problem.
 
.. math hurts my brain >.> but i agree with zyflair here as 98% just seems way to high of a chance...
 
Here's something to chew on.
If out of 52 cards your odds are 98% to draw Candy+Stage 2...then why aren't your odds so high when you draw your initial starting hand of 7? There you draw 7 out of 60, here you draw 6 out of 52. Iunno...
 
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